P4389 付公主的揹包
阿新 • • 發佈:2018-12-29
還是搞不明白生成函式是什麼東西……
首先設對於體積為\(v\)的物品,它的生成函式為\(f(x)=\sum_{i\geq 0} x^{vi}\),那麼答案的生成函式就是所有的物品的生成函式的乘積,複雜度為\(O(nm\log n)\)
於是考慮把所有生成函式取\(\ln\)相加再\(\exp\)回去,設\(g(x)=\ln(f(x))\),因為有\(f(x)=\frac{1}{1-x^v}\)
所以\[g'(x)=\frac{f'(x)}{f(x)}\]
\[g'(x)=(1-x^v)f'(x)\]
\[g'(x)=(1-x^v)\sum_{i\geq 0}vix^{vi-1}\]
\[g'(x)=\sum_{i\geq0}v(i-(i-1))x^{vi-1}\]
\[g'(x)=\sum_{i\geq0}vx^{vi-1}\]
然後積分\[g(x)=\sum_{i \geq 1} \frac{1}{i}x^{vi}\]
然後用多項式\(\exp\)求出\(f\)就可以了
//minamoto #include<bits/stdc++.h> #define R register #define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i) #define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i) #define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v) using namespace std; char buf[1<<21],*p1=buf,*p2=buf; inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;} int read(){ R int res,f=1;R char ch; while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1); for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0'); return res*f; } char sr[1<<21],z[20];int K=-1,Z=0; inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;} void print(R int x){ if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x; while(z[++Z]=x%10+48,x/=10); while(sr[++K]=z[Z],--Z);sr[++K]='\n'; } const int N=5e5+5,P=998244353,Gi=332748118; inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;} inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;} inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;} int ksm(R int x,R int y){ R int res=1; for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x); return res; } int r[N],O[N],A[N],B[N],C[N],D[N],F[N],G[N],H[N]; int n,m,x,cnt[N]; void NTT(int *A,int ty,int len){ int lim=1,l=0;while(lim<len)lim<<=1,++l; fp(i,0,lim-1)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1)); fp(i,0,lim-1)if(i<r[i])swap(A[i],A[r[i]]); for(R int mid=1;mid<lim;mid<<=1){ int I=(mid<<1),Wn=ksm(ty==1?3:Gi,(P-1)/I);O[0]=1; fp(i,1,mid-1)O[i]=mul(O[i-1],Wn); for(R int j=0;j<lim;j+=I)for(R int k=0;k<mid;++k){ int x=A[j+k],y=mul(O[k],A[j+k+mid]); A[j+k]=add(x,y),A[j+k+mid]=dec(x,y); } }if(ty==-1)for(R int i=0,inv=ksm(lim,P-2);i<lim;++i)A[i]=mul(A[i],inv); } void Inv(int *a,int *b,int len){ if(len==1)return (void)(b[0]=ksm(a[0],P-2)); Inv(a,b,len>>1);fp(i,0,len-1)A[i]=a[i],B[i]=b[i]; NTT(A,1,len<<1),NTT(B,1,len<<1); fp(i,0,(len<<1)-1)A[i]=mul(A[i],mul(B[i],B[i])); NTT(A,-1,len<<1);fp(i,0,len-1)b[i]=dec(mul(b[i],2),A[i]); fp(i,0,(len<<1)-1)A[i]=B[i]=0; } void Direv(int *a,int *b,int len){ fp(i,1,len-1)b[i-1]=mul(a[i],i);b[len-1]=0; } void Inter(int *a,int *b,int len){ fp(i,1,len-1)b[i]=mul(a[i-1],ksm(i,P-2));b[0]=0; } void Ln(int *a,int *b,int len){ Direv(a,C,len),Inv(a,D,len); NTT(C,1,len<<1),NTT(D,1,len<<1); fp(i,0,(len<<1)-1)C[i]=mul(C[i],D[i]); NTT(C,-1,len<<1),Inter(C,b,len); fp(i,0,(len<<1)-1)C[i]=D[i]=0; } void Exp(int *a,int *b,int len){ if(len==1)return (void)(b[0]=1); Exp(a,b,len>>1),Ln(b,F,len); F[0]=dec(a[0]+1,F[0]);fp(i,1,len-1)F[i]=dec(a[i],F[i]); NTT(F,1,len<<1),NTT(b,1,len<<1); fp(i,0,(len<<1)-1)b[i]=mul(b[i],F[i]); NTT(b,-1,len<<1);fp(i,len,(len<<1)-1)b[i]=F[i]=0; } int main(){ // freopen("testdata.in","r",stdin); n=read(),m=read(); fp(i,1,n)x=read(),++cnt[x]; fp(i,1,m)if(cnt[i]){ for(R int j=i;j<=m;j+=i) G[j]=add(G[j],mul(cnt[i],mul(i,ksm(j,P-2)))); }int len=1;while(len<=m)len<<=1; Exp(G,H,len);fp(i,1,m)print(H[i]); return Ot(),0; }